Polynomial Functions

This may look intimidating... but the definition of a polynomial function is:

ƒ(x) = a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2+… a₂x²+a₁x+a₀

• The value of n must be a non-negative integer (i.e. it must be a whole number or equal tozero)
• The coefficients are a_n, a_n-1…a₁, a_0.
• The degree of the polynomial function is the highest value for n, where a_n is not zero.

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Polynomials have names according to the highest degree:

When you want to find the axis of symmetry, use the formula
As a side note, it is important to remember that
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• The vertex form is ƒ(x) = a(x-h)²+k

You can use the Algebra 2 concept of "Completing the Square" to derive a vertex formula from a quadratic formula.

Example:

ƒ(x) = x²-6x+5

use

to find the "c" term you must add to the quadratic in order to make it a perfect square. Remember to also add or subtract this number from outside the parenthesis.

= (x²-6x+9)+5-9

=(x-3) ²-4

ƒ(x)=(x-3)2-4

If you want to check your answer, graph them both on your graphing calculator and you should get the same graph.

If you are given a vertex point of (3, -4) you can now write the equation for it: ƒ(x) = (x-3)²-4.

Example of finding the vertex from an equation:

ƒ(x) = 2x²-8x+3

=2(x²-4x+4)+8+3

=-2(x+2) ²+11

vertex is (-2, 11)

How to find an equation form a vertex:

• You are given a vertex point of (5, 2).
  

• From this you can figure that the equation is ƒ(x)=a(x-5) ²+2

• And if told another point on the line, you can plug x and y in the equation to find a.

How to find the vertex from the x-intercepts:

• You are given the x-intercepts of -1 and 3.
• You derive the equation y= a(x+1)(x-3) from these intercepts.
• Change the formula to vertex form
• Plug in one of the intercepts (0, -1)(0, 3) in for x and y, in order to find a.